Verified Solution • 2025–26 Session moderate

QJEEM17Mac0b4: JEE Main 2017 Mathematics - Class 12 JEEM17Mac0b4 Solution: If    then k is equal to :

If    ∫12dx(x2−2x+4)32=kk+5,\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = {k \over {k + 5}}, then k is equal to :

Multiple Choice Options

  • A 1
  • B 2
  • C 3
  • D 4

Step-by-Step Verified Solution

Given, I = ∫12dx(x2−2x+4)32\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}}

= ∫12dx[(x−1)2+3]32\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}}

Let x −- 1 = 3\sqrt 3 tan θ\theta

⇒\Rightarrow    \,\,\, dx = 3\sqrt 3 sec2 θ\theta d θ\theta

When x = 1, then θ\theta = 0

and when x = 2, θ\theta = π6{\pi \over 6}

I = ∫0π63sec⁡2θ dθ(3tan⁡2θ+3)32\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}}

= ∫0π63sec⁡2θ dθ(3sec⁡2θ)32\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}}

= ∫0π63sec⁡2θ dθ33sec⁡3θ\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}}

= 13{1 \over 3} ∫0π6dθsec⁡θ\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}}

= 13∫0π6cos⁡θ dθ{1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta }

= 13[sin⁡θ]θπ6{1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}

= 13×12{1 \over 3} \times {1 \over 2}

= 16{1 \over 6}

∴   \therefore\,\,\, According to the question,

kk+5{k \over {k + 5}} = 16{1 \over 6}

⇒\Rightarrow    \,\,\, 6k = k + 5

⇒\Rightarrow    \,\,\, k = 1

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